Counting principles, partial fractions, systems of equations – IB Maths exam questions with mark scheme

1CoreNon-calculatorHL only4 marks

A club has 8 members.

(a) A committee of 3 members is chosen. Find the number of possible committees. [2]

(b) Instead, a president, a vice-president and a secretary are chosen from the 8 members. Find the number of ways this can be done. [2]

$\binom83$ (or $\dfrac{8\times7\times6}{3!}$)M1
$56$A1
$8\times7\times6$ or $^8P_3$M1
$336$A1
Answer: (a) $56$ (b) $336$
Examiner tip: Order matters in (b) but not in (a): that is why (b) is $3!$ times larger.

2StandardCalculatorHL only6 marks

Consider the 10 letters of the word STATISTICS.

(a) Find the number of different arrangements of all 10 letters. [3]

(b) Find the number of these arrangements in which the three letters S are next to each other. [3]

$\dfrac{10!}{\cdots}$ (correct numerator)M1
Divides by $3!\,3!\,2!$ for repeated S, T, IM1
$50\,400$A1
Treats SSS as one object, giving 8 objectsM1
$\dfrac{8!}{3!\,2!}$ (repeats of T and I)M1
$3360$A1
Answer: (a) $50\,400$ (b) $3360$
Examiner tip: Divide by the factorial of every repeated letter, including those in the block method.

3StandardNon-calculatorHL only4 marks

A team of 4 is chosen from 6 men and 5 women. The team must contain at least 2 women. Find the number of possible teams.

Valid method: cases (2, 3, 4 women) or total minus (0 or 1 woman)M1
Complement: $\binom{11}4=330$ and $\binom64=15$ (no women); or direct: $\binom52\binom62=150$ (2 women)A1
Complement: $\binom51\binom63=100$ (one woman); or direct: $\binom53\binom61=60$ and $\binom54=5$A1
$330-15-100=215$ or $150+60+5=215$A1
Answer: $215$
Examiner tip: 'At least' questions are often quicker by complement; check by the direct method if time allows.

4HardNon-calculatorHL only4 marks

How many four-digit numbers greater than 4000 can be formed from the digits $1,2,3,4,5,6,7$ with no digit used more than once, if the number must be even?

Splits into cases by parity of the first digitM1
First digit even (4 or 6): $2\times2\times5\times4=80$A1
First digit odd (5 or 7): $2\times3\times5\times4=120$A1
$200$A1
Answer: $200$
Examiner tip: Fill the most restricted positions (first and last digits) first, and watch the overlap when both must be even.

5StandardNon-calculatorHL only4 marks

Write $\dfrac{5x+1}{(x-1)(x+2)}$ in the form $\dfrac{A}{x-1}+\dfrac{B}{x+2}$, where $A,B\in\mathbb{Z}$.

$5x+1=A(x+2)+B(x-1)$M1
Substitutes $x=1$ or $x=-2$ (or equates coefficients)M1
$A=2$A1
$B=3$A1
Answer: $\dfrac2{x-1}+\dfrac3{x+2}$
Examiner tip: Check by substituting a third value of $x$, e.g. $x=0$: $\tfrac{1}{-2}=-2+\tfrac32$.

6HardNon-calculatorHL only6 marks

(a) Express $\dfrac{x+7}{(x-1)(x+3)}$ in partial fractions. [3]

(b) Hence find the exact value of $\displaystyle\int_2^5\frac{x+7}{(x-1)(x+3)}\,\mathrm dx$, giving your answer in the form $\ln k$, where $k\in\mathbb{Z}$. [3]

$x+7=A(x+3)+B(x-1)$ (or equivalent, e.g. cover-up method)M1
$A=2$A1
$B=-1$, giving $\dfrac2{x-1}-\dfrac1{x+3}$A1
Integrates both terms to logarithmsM1
$2\ln|x-1|-\ln|x+3|$ (allow without modulus signs; ignore $+c$)A1
$(2\ln4-\ln8)-(2\ln1-\ln5)=\ln\dfrac{16\times5}{8}=\ln10$A1
Answer: (a) $\dfrac2{x-1}-\dfrac1{x+3}$ (b) $\ln10$
Examiner tip: Keep the coefficient 2 with its log until the end, then use $2\ln4=\ln16$ and combine the logs into a single $\ln k$.

7HardNon-calculatorHL only6 marks

Consider the system of equations, where $m\in\mathbb{R}$:

$$x+y+z=1,\quad x+2y+3z=4,\quad 2x+3y+4z=m.$$

(a) Show that the system has solutions only when $m=5$. [4]

(b) For $m=5$, find the general solution. [2]

Eliminates $x$ from equations 1 and 2: $y+2z=3$M1
Eliminates $x$ from equations 1 and 3: $y+2z=m-2$M1
The two results require $m-2=3$A1
If $m\ne5$ the equations are inconsistent (no solution), so solutions exist only for $m=5$R1
Sets $z=t$: $y=3-2t$M1
$x=t-2$, so $(x,y,z)=(t-2,\ 3-2t,\ t)$A1
Answer: (a) shown ($m=5$) (b) $(x,y,z)=(t-2,\,3-2t,\,t)$
Examiner tip: Row 3 is row 1 + row 2 on the left, which is why the right-hand sides must agree: 1 + 4 = 5.

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